Please correct answer and don't use hand rating Calculate the pH of a 0.154 M solution of ethylenediamine (H,NCH2CH2NH2). The pKa values for the acidic form ofthylenediamine (H3NCH2CH2NH) are 6.848 (pKa1) and 9.928 (pKa2).pH =11.51Correct AnswerCalculate the concentration of each form of ethylenediamine in this solution at equilibrium.[H2NCH2CH2NH] =4.12 x10-3Incorrect Answer7.05 x10-8[HNCH2CH2NH] =Correct AnswerMM

Step one:The pH indicated is 11.51 and the pKa which applies to the second dissociation of…

Answered: Calculate the pH of a 0.154 M solution…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter15: Complex Acid/base Systems

Section: Chapter Questions

Problem 15.16QAP

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Acid-Base Equilibria Many factors contribute to the acidity of organic compounds. Electronegativity, resonance, induction, hybridization, aromaticity, and atomic size, all play a role. In the following comparisons, you are asked to identify the factor(s) that would be most important to analyze when predicting relative acidity, and then to predict the trend in acidity and pKa values. For each of the following pairs of compounds answer the following two multiple-choice questions. 1. What factor(s) are the most important to consider when predicting the relative acidity of the two compounds? a. Electronegativity of the atom possessing the hydrogen. b. Resonance stabilization of the anionic conjugate base. c. Inductive stabilization of the anionic conjugate base. d. Hybridization of the atom possessing the hydrogen. e. The atomic size of the atom possessing the hydrogen.
Calculate the pH of a 0.123 M solution of ethylenediamine (H₂NCH₂CH₂NH₂). The pKą values for the acidic form of ethylenediamine (H3NCH₂CH₂NH³) are 6.848 (pKal) and 9.928 (pKa2). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H₂NCH₂CH₂NH₂] : = [H₂NCH₂CH₂NH3] = M M
Calculate the pH of a 0.295 M solution of ethylenediamine (H₂NCH,CH,NH₂). The pKa values for the acidic form of ethylenediamine (HNCH,CH,NH) are 6.848 (pKa1) and 9.928 (pKa2). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H2NCH2CH2NH₂] = [H2NCH,CH,NH] = [HNCH,CH,NH3] = M M M
Calculate the pH of a 0.172 M solution of ethylenediamine (H,NCH,CH,NH,). The pK, values for the acidic form of ethylenediamine (HNCH,CH,NH) are 6.848 (pKat) and 9.928 (pKaz). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H₂NCH,CH,NH₂] [H2NCH2CH2NH] = [HNCH2CH2NH] = M M M
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Calculate the pH of a 0.365 M solution of ethylenediamine (H,NCH, CH, NH,). The pKa values for the acidic form of ethylenediamine (H†NCH,CH,NH†) are 6.848 (pKa1) and 9.928 (pKa2). pH =| 11.25 Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H,NCH,CH,NH,] = M [H,NCH,CH,NH)= M [HNCH,CH,NH;) = M
Calculate the pH at 25°C of a 0.64M solution of anilinium chloride C6H5NH3Cl. Note that aniline C6H5NH2 is a weak base with a pKb of 4.87.
Calculate the pH at 25°C of a 0.89M solution of trimethylammonium chloride CH33NHCl. Note that trimethylamine CH33N is a weak base with a pKb of 4.19 .Round your answer to 1 decimal place.
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The acid dissociation constant K of trimethylacetic acid (HC(CH³)₂CO₂) is 9.33 × 10¯6. a Calculate the pH of a 3.6M solution of trimethylacetic acid. Round your answer to 1 decimal place. pH = - 0 X Ś
The pH at 25 °C of an aqueous solution of the sodium salt of α-naphthol (NaC10H7O) is 11.23. Calculate the concentration of C10H7O- in this solution, in moles per liter. Ka for HC10H7O is equal to 4.6×10-10.

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